Why is armv6 a Valid Architecture for iOS 5?

Question

Can someone tell me why creating a new project with an iOS Deployment Target of iOS 5.0 includes armv6 as a Valid Architecture in Project > Build Settings by default? The default info.plist file only lists armv7 in the Required Device Capabilities.

As I understand it, if I am building for iOS 5 or later that will only run on armv7 devices (i.e. iPhone 3GS or later, iPod 3G or later, and the iPads). Thus I should be able to remove all the armv6 references, and set Build Active Architecture Only to YES.

Can anyone confirm or clarify? Thanks.

XCode v4.3.2, Base SDK 5.1, iOS Deployment Target iOS 5.0, iPhone Device Family

PS - I understand that removing armv6 will reduce the size of the binary.

Answer 1

$(VALID_ARCHS) defines the set of architectures that an Xcode target supports. $(ARCHS) tells Xcode what architectures to build. What actually gets built for each target is the intersection of $(VALID_ARCHS) and $(ARCHS).

$(VALID_ARCHS) is typically used to avoid building a target for some architecture that the project as a whole otherwise supports. If you don't need to do that, you can safely ignore it.